"Étalification" of a scheme

Question

Let $X$ be a scheme. Does the forgetful functor

$$\mathrm{EtSch}/X \to \mathrm{Sch}/X$$

have a right adjoint $Z \mapsto \tilde{Z}$? One might call $\tilde{Z}$ the étalification of $Z$. So this is an étale $X$-scheme together with an $X$-morphism $\tilde{Z} \to Z$, which induces for every étale $X$-scheme $Y$ a bijection $\hom_X(Y,\tilde{Z}) \cong \hom_X(Y,Z)$.

If $X$ is the spectrum of a field $k$, the answer is yes, using 1) the equivalence of sites between étale $k$-schemes and continuous $G$-sets, where $G$ is the absolute Galois group of $k$, 2) the fact that every sheaf on $G$-sets is representable. Explicitly: If $Z$ is a $k$-scheme, let us denote by $Z_{\mathrm{sep}}$ the subset of all points $z \in Z$ such that $k(z)/k$ is finite and separable. Then $$\tilde{Z} = \coprod\limits_{z \in Z_{\mathrm{sep}}} \mathrm{Spec}(k(z)).$$ For example, $\widetilde{\mathbb{A}^1}=\coprod_{\alpha \in k} \mathrm{Spec}(k)$ and $\widetilde{\mathbb{A}^2}$ is the coproduct of spectra of the form $(k[x]/(p)[T])/(q)$, where $p \in k[x]$ is irreducible and $q$ is some irreducible separable polynomial over $k[x]/(p)$.

Perhaps this construction is well-known, therefore I've put the reference request tag.

Answer 1

Given a scheme $Z$ over $X$, one can consider the sheaf of "sections $X\to Z$". More precisely, that sheaf associates to every étale open $U\to X$ the set of commutative diagrams $$ \begin{matrix} & & Z \\\ & \nearrow & \downarrow \\\ U & \to & X \end{matrix} $$ The scheme $\tilde Z$ you are looking for is the espace étalé of the above sheaf, a highly non-separated scheme.

Answer 2

Can we just do this the hard way?

First we construct the set of points of the etalification $\bar{Z}$. A point of the etalification is a point $P$ of $X$ plus a map from the spectrum $T$ of the etale local ring at $X$ to $Z$ that forms a commutative diagram with the natural map $T \to X$. (or, if $Z \to X$ is not finite type, we need to use a finitely-generated subring of the etale local ring.)

Then for $Y\to X$ is etale with a map $Y \to Z$. We get a set-theoretic map $Y \to \bar{Z}$. One easily checks that each point is in the image of such a map. We form the finest topology such that the image of each such map is open.

Now we need a sheaf of rings. The natural map $\bar{Z} \to Z$ given by taking the image of the special fiber is continuous. Pull back the structure sheaf of $Z$, mod out by the functions on a set $U$ that are $0$ in the images of the open balls $T$ corresponding to the points of $U$, and sheafify.

We will check that this is a scheme by finding affine neighborhoods of each point. Each point of $\bar{Z}$ is in the image of some etale $Y\to X$, and we can further choose $Y$ to be affome amd $T \to Z$ an immersion. We will check that $T \to \bar{Z}$ is an inclusion of locally-ringed spaces. Clearly it is injective. Next we check that it is continuous and open. Open subsets of $\bar{Z}$ wil come from other etale maps which agree on some etale local ring, but then these must agree on an open subset, so they will give open subsets of $Y$. Open subsets of $Y$ are just more schemes etale over $X$.

Then we check it gives an isomorphism of sheaves of rings. The sheaf of rings on $Y$ is the pullback of the sheaf of rings of $Z$, mod the functions that vanish on it. But the functions that vanish on $Y$ are exactly the functions that vanish on each etale local rings, so the two sheaves of rings are identical.

So in addition $\bar{Z}$ is etale over $X$. By construction of the sheaf of rings, $\bar{Z} \to Z$ is a morphism. Similarly the set-theoretic maps $Y \to \bar{Z}$ for all $Y\to Z$ etale over $X$ are clearly morphisms, and give the required bijections.

Obviously this is quite sketchy and there are a lot of technical details I'm missing but it seems doable.